3.240 \(\int \frac{(a+b \sin ^{-1}(c x))^2}{x \sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=257 \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d-c^2 d x^2}} \]

[Out]

(-2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] + ((2*I)*b*Sqrt[1
- c^2*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] - ((2*I)*b*Sqrt[1 - c^2*x^2
]*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] - (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3,
-E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x])])/Sqrt[d - c
^2*d*x^2]

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Rubi [A]  time = 0.340697, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4713, 4709, 4183, 2531, 2282, 6589} \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^2/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

(-2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] + ((2*I)*b*Sqrt[1
- c^2*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] - ((2*I)*b*Sqrt[1 - c^2*x^2
]*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] - (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3,
-E^(I*ArcSin[c*x])])/Sqrt[d - c^2*d*x^2] + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x])])/Sqrt[d - c
^2*d*x^2]

Rule 4713

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[
Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rule 4709

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt{d-c^2 d x^2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \sqrt{1-c^2 x^2}} \, dx}{\sqrt{d-c^2 d x^2}}\\ &=\frac{\sqrt{1-c^2 x^2} \operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{\left (2 b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}+\frac{\left (2 b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{\left (2 i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}+\frac{\left (2 i b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{\left (2 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{\left (2 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{Li}_3\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{Li}_3\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.613659, size = 301, normalized size = 1.17 \[ \frac{2 a b \sqrt{1-c^2 x^2} \left (i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )}{\sqrt{d-c^2 d x^2}}+\frac{b^2 \sqrt{1-c^2 x^2} \left (2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-2 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+2 \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )}{\sqrt{d-c^2 d x^2}}-\frac{a^2 \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )}{\sqrt{d}}+\frac{a^2 \log (c x)}{\sqrt{d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])^2/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

(a^2*Log[c*x])/Sqrt[d] - (a^2*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]])/Sqrt[d] + (2*a*b*Sqrt[1 - c^2*x^2]*(ArcSin
[c*x]*(Log[1 - E^(I*ArcSin[c*x])] - Log[1 + E^(I*ArcSin[c*x])]) + I*PolyLog[2, -E^(I*ArcSin[c*x])] - I*PolyLog
[2, E^(I*ArcSin[c*x])]))/Sqrt[d - c^2*d*x^2] + (b^2*Sqrt[1 - c^2*x^2]*(ArcSin[c*x]^2*Log[1 - E^(I*ArcSin[c*x])
] - ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] + (2*I)*ArcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)*ArcSin
[c*x]*PolyLog[2, E^(I*ArcSin[c*x])] - 2*PolyLog[3, -E^(I*ArcSin[c*x])] + 2*PolyLog[3, E^(I*ArcSin[c*x])]))/Sqr
t[d - c^2*d*x^2]

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Maple [A]  time = 0.152, size = 387, normalized size = 1.5 \begin{align*} -{{a}^{2}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{-{c}^{2}d{x}^{2}+d} \right ) } \right ){\frac{1}{\sqrt{d}}}}+{\frac{{b}^{2}}{d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) - \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,i\arcsin \left ( cx \right ){\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,i\arcsin \left ( cx \right ){\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,{\it polylog} \left ( 3,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,{\it polylog} \left ( 3,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) }-{\frac{2\,iab}{d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( i\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -i\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/x/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-a^2/d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)+b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*(arcsin(
c*x)^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-arcsin(c*x)^2*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-2*I*arcsin(c*x)*polylog(2,-
I*c*x-(-c^2*x^2+1)^(1/2))+2*I*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*polylog(3,-I*c*x-(-c^2*x^2+1)^
(1/2))-2*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2)))/d/(c^2*x^2-1)-2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*
(I*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-I*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+polylog(2,-I*c*x-(-
c^2*x^2+1)^(1/2))-polylog(2,I*c*x+(-c^2*x^2+1)^(1/2)))/d/(c^2*x^2-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{2} d x^{3} - d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^2*d*x^3 - d*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}}{x \sqrt{- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**2/x/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))**2/(x*sqrt(-d*(c*x - 1)*(c*x + 1))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt{-c^{2} d x^{2} + d} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/x/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2/(sqrt(-c^2*d*x^2 + d)*x), x)